使用PHP将数据从MYSQL转换为JSON [英] Getting data from MYSQL into JSON using PHP

问题描述

我有以下非常简单的测试PHP代码，该代码提取数据并将其放入JSON格式的文本中.

I have the following quite simple test PHP code that extracts the data and puts it into JSON formatted text.

我收到以下错误.

致命错误:在第33行的/var/www/test.php中，允许的内存大小为33554432字节已用尽(试图分配1979603字节)

Fatal error: Allowed memory size of 33554432 bytes exhausted (tried to allocate 1979603 bytes) in /var/www/test.php on line 33

第33行是json_encode()行.

有没有办法使它更有效? PHP.ini已被设置为最大32M，因此从8M标准开始可以调整大小！

Is there a way to make this more efficient? The PHP.ini is already set to 32M as max, hence sized up from the 8M standard!

 <?php
    require('../../admin/db_login.php');

    $db=mysql_connect($host, $username, $password) or die('Could not connect');
    mysql_select_db($db_name, $db) or die('');

    $result = mysql_query("SELECT * from listinfo") or die('Could not query');
    $json = array();

    if(mysql_num_rows($result)){
            $row=mysql_fetch_assoc($result);
        while($row=mysql_fetch_row($result)){
            //  cast results to specific data types

            $test_data[]=$row;
        }
        $json['testData']=$test_data;
    }

    mysql_close($db);

    echo json_encode($json);


    ?>

推荐答案

您可能正在编码非常大的数据集.您可以对每一行进行编码，而不是一次大的操作即可对每一行进行编码.

You are probably encoding a very large dataset. You could encode each row, one row at a time instead of encoding it in one big operation.

<?php
require('../../admin/db_login.php');

$db=mysql_connect($host, $username, $password) or die('Could not connect');
mysql_select_db($db_name, $db) or die('');

$result = mysql_query("SELECT * from listinfo") or die('Could not query');

if(mysql_num_rows($result)){
    echo '{"testData":[';

    $first = true;
    $row=mysql_fetch_assoc($result);
    while($row=mysql_fetch_row($result)){
        //  cast results to specific data types

        if($first) {
            $first = false;
        } else {
            echo ',';
        }
        echo json_encode($row);
    }
    echo ']}';
} else {
    echo '[]';
}

mysql_close($db);

这样，每次对json_encode()的调用仅编码一个小数组，而不是一个大数组.最终结果是相同的. 这是IMO解决方案，它将使用较少的内存.

That way, each call to json_encode() only encodes a small array instead of a large one. The end result is the same. This is IMO the solution which will use the less memory.

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